Answer:
a = 2.85714 m/s^2
Explanation:
Fnet=ma
eaqual and oppisate forces means that the 100N pushed on the ball comes back to the child.
100 = 35*a
a= 100/35
a force of 193 pounds makes an angle of 86.27 with a second force. The resultant of the two forces makes an angle of 31.4 to the first force. Find the magnitudes of the second force and of the resultant.
Answer:
Explanation:
Let force P = 193 pounds
second force = Q
Angle that resultant makes with force P is θ
Tanθ = Q sin86.27 / (P + Q cos86.27 )
Tan 31.4 = Q sin86.27 / (193 + Q cos86.27 )
.61 = .99 Q / (193 + .065Q )
117.73 + .04 Q = .99Q
117.73 = .95 Q
Q = 123.93 pounds .
A mass that has a force of the 10N to the right, 7N up, 3N down, and 4N to the left. What is the net external force on that mass? If the mass is 2kg find the acceleration.
The gravity on the moon is 1/6 as strong as that of earth.An object on earth weighs 1 Newton another object on the moon also weighs 1 Newton.Which of the two objects has the greater mass
Answer:
The object on the moon has greater mass
Explanation:
Given;
weight of the object on Earth, W₁ = 1 N
wight of the object on Moon, W₂ = 1 N
gravity on Earth, g₁ = 9.8 m/s²
gravity on the Moon, g₂ = g₁/6 = 1.633 m/s²
Apply Newton's second law of motion;
Weight of the object on Earth is given by;
W₁ = m₁g₁
m₁ = W₁/g₁
m₁ = 1 / 9.8
m₁ = 0.102 kg
Weight of the object on Moon is given by;
W₂ = m₂g₂
m₂ = W₂ / g₂
m₂ = 1 / 1.633
m₂ = 0.612 kg
Therefore, the object on the moon has greater mass.
Based on these photographs, how do Mars's surface and atmosphere differ from Earth's?
Answer:
Explanation:
Mars does not appear to have any water, while Earth is mostly water. Mars also has yellow-brown sky (pink-brown at sunrise and sunset), which suggests that it has a different atmospheric makeup.
Answer:
Mars does not appear to have any water, while Earth is mostly water. Mars also has yellow-brown sky (pink-brown at sunrise and sunset), which suggests that it has a different atmospheric makeup.
Explanation:
A uniform string of length 2.5 m and mass 0.01 kg is placed under a tension of 10 N. (5 pts) A) What is the frequency of the fundamental mode
Answer:
The fundamental frequency is 10 Hz
Explanation:
The frequency of the fundamental mode of the string is given by;
[tex]f_o =\frac{1}{2l}\sqrt{\frac{T}{\mu }[/tex]
where;
L is length of the string = 2.5 m
T is tension of the string, = 10 N
μ is mass per unit length = m / L = 0.01 / 2.5 = 0.004 kg/m
Substitute the given values and solve for the frequency;
[tex]f_o =\frac{1}{2l}\sqrt{\frac{T}{\mu }}\\\\ f_o =\frac{1}{2*2.5}\sqrt{\frac{10}{0.004}}\\\\ f_o = 0.2\sqrt{2500}\\\\ f_o = 0.2 *50\\\\f_o = 10 \ Hz[/tex]
Therefore, the fundamental frequency is 10 Hz
What change to the atomic model helped solve the problem seen in Rutherford’s model?
A segment of wire of total length 3 m carries a 10 A current and is formed into a semicircle. Determine the magnitude of the magnetic field (in micro-tesla) at the center of the circle along which the wire is placed.
Answer:
[tex]B=3.29\ \mu T[/tex]
Explanation:
Given that,
Length of a wire is 3 m
Current in the wore, I = 10 A
The wire formed a semicircle.
We need to find the magnitude of the magnetic field (in micro-tesla) at the center of the circle along which the wire is placed.
The magnetic field at the center of semicircle is given by :
[tex]B=\dfrac{\mu_o I}{4R}[/tex]
R is radius of semicircle,
[tex]R=\dfrac{3}{\pi}=0.954\ m[/tex]
So,
[tex]B=\dfrac{4\pi \times 10^{-7}\times 10I}{4(0.954)}\\\\B=3.29\times 10^{-6}\ T\\\\B=3.29\ \mu T[/tex]
So, the magnitude of the magnetic field is [tex]3.29\ \mu T[/tex].
In a class experiment to determine information about free-fall acceleration, a watermelon and a pumpkin are each set to fall from the back of the stands at your football stadium.
a. If the watermelon and the pumpkin are both dropped at the same time, which one will hit the ground first?
b. If the watermelon is thrown downward with an initial speed of 10 m/s and the pumpkin is dropped, which one will hit the ground first? Show calculation to support your answer.
c. If it takes the watermelon 1 second to reach the ground when it is thrown downward at 10 m/s, how tall are the stands?
d. How long does it take the pumpkin to reach the ground if it is dropped from this height you calculated in part c?
Answer:
a) They both hit the ground at the same time
b) Watermelon will hit first, since its speed is faster than the pumpkin speed
c) h = 29,6 m
d)t = 3,02 sec
Explanation:
Equations for fall free movement are:
vf = v₀ + g*t when v₀ = 0 (dropped case) vf = g*t
h = v₀*t + 1/2*g*t²
a) For both ( watermelon and pumpkin) the equation of speed is the same:
vf = g*t² Both will have the same speed second through second
They both hit the ground at the same time
b) Now is watermelon is thrown with v₀ = 10 m/s
Watermelon will hit first since its speed is faster than the pumpkin speed
vf(watermelon) = 10 + g*t
vf₂ (pumpkin) = g*t
c) h = v₀*t + (1/2)*g*t²
h = (10)*1 + (1/2)*9,8*1
h = 10 + 19,6
h = 29,6 m
d) h = g*t
t = 29,6/9,8
t = 3,02 sec
A 55kg skateboarder wants to just make it to the upper edge of a quarter pipe, a pipe that is one-quarter of a circle with a radius of 3.10m. What speed does he need at the bottom?
Answer:
Explanation:
A 52.0 kg skateboarder wants to just make it to the upper edge of a "quarter pipe," a track that is one-quarter of a circle with a radius of 2.60 m .
What speed does he need at the bottom
Let the lowest point of the circle = (0,0)
Center of circle = (0, 2.6)
The height of the circle = 2 * radius = 2 * 2.60 = 5.20 m
Highest point = (0, 5.2)
As the skateboarder moves around ¼ of a vertical circle, the skateboarder moves from the lowest position to a position that is ½ the way up to the highest position. This is the point that is 2.6 meters directly to the right of left of the center = (2.6, 2.6)
As the skateboarder has moved 2.6 m upward, the potential energy increase = m * g * ∆h = 52.0 * 9.8 * 2.6
During this same time, the kinetic energy decreased from the maximum to 0. The decrease of KE = the increase of PE
½ * m * ∆v^2 = m * g * ∆h
½ * 52 * ∆v^2 = 52.0 * 9.8 * 2.6
½ * ∆v^2 = 9.8 * 2.6
∆v = (2 * 9.8 * 2.6)^0.5 = 7.14 m/s
The velocity at highest point, (2.6,2.6) is 0 m/s
So the velocity at the lowest point must be 7.14 m/s
Let’s see if the skateboarder has enough KE to move upward 2.6 m.
KE at bottom = ½ * 52 * 7.14^2 = 1325.5
PE = 52 * 9.8 * h
52 * 9.8 * h = 1325.5
h = 2.6 m
The answer is OK
The speed needed by the skateboarder at the bottom is 7.8 m/s.
The given parameters;
mass of the skateboarder, m = 55 kgradius of the quarter pipe, r = 3.1 mApply the principle of conservation of mechanical energy to determine the sped of the skateboarder at the bottom;
[tex]P.E_{top} = K.E_{bottom}\\\\mgh = \frac{1}{2}mv^2\\\\gh = \frac{1}{2} v^2\\\\v^2 = 2gh\\\\v = \sqrt{2gh} \\\\v = \sqrt{2\times 9.8 \times 3.1} \\\\v = 7.8 \ m/s[/tex]
Thus, the speed needed by the skateboarder at the bottom is 7.8 m/s.
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The tendon from Lissa’s knee extensor muscles attaches to the tibia bone 1.5 in. (4 cm) below the center of her knee joint, and her foot is 15 in. (38 cm) away from her knee joint. What arc length does Lissa’s foot move through when her knee extensor muscles contract and their point of insertion on the tibia moves through an arc length of 2 in. (5 cm)?
Answer:
s₂ = 0.475 m = 47.5 cm
Explanation:
The arc length and the angle of rotation are related through the formula:
s = rθ
where,
s = arc length
r = radius of curvature
θ = angle of rotation
First, we consider the arc length covered by the point of insertion of extensor muscles.
s₁ = r₁θ
where,
s₁ = arc length covered by insertion of extensor muscle = 5 cm
r₁ = length of insertion from knee = 4 cm
θ = Angle of Rotation = ?
Therefore,
5 cm = (4 cm)(θ)
θ = (5 cm)/(4 cm)
θ = 1.25 rad
Now, we consider the arc length covered by the foot.
s₂ = r₂θ
where,
s₂ = arc length covered by the foot = ?
r₂ = distance from knee to foot = 38 cm = 0.38 m
The angle of rotation will be the same for the foot as the insertion.
Therefore,
s₂ = (0.38 m)(1.25 rad)
s₂ = 0.475 m = 47.5 cm
Which is one way that ultrasound technology may be used?
O repairing fractured bones
O locating misplaced electrical equipment
O cleaning dirty surgical tools
O canceling out unwanted noise
Answer:
Locating misplaced electrical equipment.
Explanation:
Ultrasound is primarily used by doctors to locate fractured bones or damaged organs, so I assume it can be used to locate other things too.
Find μsμsmu_s, the coefficient of static friction between the rod and the rails. Express the coefficient of static friction in terms of variables given in the introduction.
Answer:
The question is incomplete. However the complete question is :
A Rail Gun uses electromagnetic forces to accelerate a projectile to very high velocities. The basic mechanism of acceleration is relatively simple and can be illustrated in the following example. A metal rod of mass m and electrical resistance R rests on parallel horizontal rails (that have negligible electric resistance), which are a distance L apart. The rails are also connected to a voltage source V, so a current loop is formed.
The vertical magnetic field, initially zero, is slowly increased. When the field strength reaches the value [tex]$B_0$[/tex], the rod, which was initially at rest, begins to move. Assume that the rod has a slightly flattened bottom so that it slides instead of rolling. Use g for the magnitude of the acceleration due to gravity.
Find μs, the coefficient of static friction between the rod and the rails.
Express the coefficient of static friction in terms of variables given in the introduction.
So the answer is : [tex]$\mu_s=\frac{\Delta V LB_0}{Rmg}$[/tex]
Explanation:
It is given : B = [tex]$B_0$[/tex]
Therefore, using force on the current carrying wire is:
[tex]$\mu_s mg = B_0IL$[/tex]
[tex]$\mu_s= \frac{ILB_0}{mg}$[/tex]
Therefore,
[tex]$\mu_s=\frac{\Delta V LB_0}{Rmg}$[/tex]
The coefficient of static friction in terms of variables will be [tex]\rm \mu_s =\frac{ \triangle VLB_0}{Rmg} \\\\[/tex].
What is the cofficient of static friction?The ratio of the greatest static friction force (F) between the surfaces in contact before movement begins to the normal (N) force is the coefficient of static friction.
The given data in the problem is;
[tex]\rm B=B_0[/tex]
The force on a current-carrying wire is balanced by the friction force.
The force on a current-carrying wire is given by;
[tex]\rm \mu_s mg= B_0IL \\\\ \rm \mu_s=\frac{ILB_0}{mg}\\\\ \rm \mu_s= \frac{\triangle VLB_0}{Rmg}[/tex]
Hence the coefficient of static friction in terms of variables will be [tex]\rm \mu_s =\frac{ \triangle VLB_0}{Rmg} \\\\[/tex].
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Two identical objects of the same size and shape are dropped. One is dropped from a table, while the other is dropped from the top of a tall building. Which will hit the ground with more impact?
Answer:
These two objects hit the ground at the same time. But their impact velocity will be different
because they are dropped from different heights.
The one that is dropped from the building will have more impact when it hits the ground because gravity causes this onject to fall toward the ground at a faster and faster velocity the longer the object falls. In fact, its velocity increases by 9.8 m/s2.
Explanation:
Initial velocity of the object u=0
Final velocity of the object when it reaches the ground v=A m/s
Using v^2−u^2=2gS where g=10m/s^2
A^2 −0^2=2(10)H
A^2=20*H H- heigh from the table
Object falling from the building will have the same initial velocity but different heigh.
T>H
so
B^2=20*T
20*T>20*H
so B^2>A^2
it equal to B>A so object falling from the building will have more impact than falling from the table.
The speed of light in vacuum is exactly 299,792,458 m/s. A beam of light has a wavelength of 621 nm in vacuum. This light propagates in a liquid whose index of refraction at this wavelength is 1.18.What is the speed of this light in the liquid
Answer:
2.54*10^8 m/s
Explanation:
Given that
Speed of light in space, c = 299792458 m/s
Wavelength of the beam of light, λ = 621 nm or 621*10^-9 m
Index of refraction of the liquid, n = 1.18
Speed of light in the liquid, v = ?
The index of refraction of an optical material, n is given by the following formula.
n = c/v, and if we substitute directly, we have
1.18 = 299792458 / v, if we make v subject of formula, we have
v = 299792458 / 1.18
v = 2.54*10^8 m/s
Which muscles pump blood through your heart and body?
a
Cardiac
b
Skeletal
c
Smooth
d
Strong
Gray used a pulley to lift a 300 N object a distance of 3 m. It took Gray 30 seconds to lift the object. How much WORK did Gray exert?
Answer:
900J
Explanation:
The Work done can be calculated using
Work done = force × distance
According to this question, force = 300N, distance = 3m
Work done = 300 × 3
Work done = 900J
Gray exerted 900J of work
What is the average velocity of the cart? Is the velocity increasing decreasing or constant as the cart moves along the air track?
Answer:
It is increasing .60 m/s
Explanation:
Two points are on a disk that rotates about an axis perpendicular to the plane of the disk at its center. Point B is 3 times as far from the axis as point A. If the linear speed of point B is V, then the linear speed of point A is:_______
a. 9V.
b. 3V.
c. V.
d. V/3.
e. V/9.
Answer:
D. V/3.
Explanation:
V = rw
At point a
Linear velocity
= Va = rw
At point b
We have,
V = 3rw
So linear velocity of a
= V/Va = 3rw/re
I cross multiples and after which I got Va = v/3
Please view attachment
Answer:
The linear speed of point A is [tex]\frac{V}{3}[/tex] (Option D)Explanation:
The relation between linear and angular velocity,
[tex]V = rw[/tex]
Now, for point A the linear velocity is
[tex]V_A = rw[/tex]
And for point B
[tex]V = 3rw[/tex]
Therefore, the linear velocity of the point A
[tex]\frac{V}{V_A} = \frac{3rw}{rw}\\\\V_A = \frac{V}{3}[/tex]
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In an old-fashioned television tube, an electron () starting from rest experiences a force of 4.0 × 10–15 N over a distance of 50 cm. Ignoring the relativistic effects, the final speed of the electron is:
Answer:
The final speed of the electron is 6.626 x 10⁷ m/s
Explanation:
force applied to the electron, f = 4 x 10⁻¹⁵ N
distance traveled by the electron, d = 50 cm = 0.5 m
The work done on the electron;
W = F x d
W = (4 x 10⁻¹⁵ N) x (0.5 m)
W = 2 x 10⁻¹⁵ J
The kinetic energy of the electron is given by;
[tex]K.E = \frac{1}{2}m_ev^2[/tex]
Apply work - energy theorem;
K.E = W
[tex]\frac{1}{2}m_ev^2 = 2 *10^{-15}\\\\ v^2 = \frac{2( 2 *10^{-15})}{m_e}\\\\v= \sqrt{\frac{2( 2 *10^{-15})}{m_e}}\\\\ v = \sqrt{\frac{2( 2 *10^{-15})}{9.11*10^{-31}}}\\\\v = 6.626*10^7 \ m/s[/tex]
Therefore, the final speed of the electron is 6.626 x 10⁷ m/s
A heating element is made by maintaining a potential difference of 75.0 V across the length of a Nichrome wire that has a 3.73 × 10–6 m2 cross section. Nichrome has a resistivity of 5.00 × 10–7 Ω·m. (a) If the element dissipates 7880 W, what is its length? (b) If 107 V is used to obtain the same dissipation rate, what should the length be?
Answer:
[tex]P=V^2/R[/tex]
[tex]R=V^2/P[/tex]
Upon substituting the values
a) R= 75*75/7800 = 0.72ohm
Also
R=ρ*L/A
L=R*A/ρ
L=[tex]\frac{0.72*3.73*10^-6 }{5*10^-7}[/tex] = 5.44m
b) when the voltage 107 is used
R=107*107/7800 = 1.468OHM
L=[tex]\frac{1.468*3.73*10^-6 }{5*10^-7}[/tex] =11.1m
If you pour liquid into a tall, narrow glass, you may hear sound with a steadily rising pitch. What is the source of the sound
Answer:
the vibration of the glass tube creating sound waves within itself. reason is the rising pitch is that the liquid rise shortens the length of the vibrating area within the tube.
Explanation:
A 75.0 kg skier slides down a 75.0 m high slope without friction. The velocity of the skier at the bottom of the slope is
Answer:
v = 38.34 m
Explanation:
Given that,
Mass of a skier, m = 75 kg
It falls down a 75.0 m high slope without friction, h = 75 m
We need to find the velocity of the skier at the bottom of the slope. It is based on the concept of conservation of energy. So, v is the required velocity. So,
[tex]mgh=\dfrac{1}{2}mv^2\\\\v=\sqrt{2gh} \\\\v=\sqrt{2\times 9.8\times 75} \\\\v=38.34\ m/s[/tex]
So, the velocity of the skier at the bottom of the slope is 38.34 m.
A 3.05 kg block is sliding across level ground and feels a 5.27 N friction force. What is the coefficient of friction?
Answer:
0.176
Explanation:
(5.27)/(3.05*9.8)
It is indeed 0.176 for the answer.
3. A 0.14-kg ball traveling with a speed of 43 m/s is brought to rest in a catcher's mitt.
Determine the magnitude of the impulse exerted by the mitt on the ball.
Answer:
change of momentum
= mass * change in velocity
= 0.14 * (36 - 0)
= 5.04 kg m/s ---answer
Explanation:
A scout hikes 2.35KM east of a campsite he takes a break for lunch and then hikes another 1.25 KM north of the location where he ate lunch. what distance is the scout from the campsite
Answer:
mk7000000.0000000000
Calculate the peak voltage (in Volts) of a generator that rotates its 200-turn, 0.100 m diameter coil at 3.5 x 103 rpm in a 0.7 T field. You should round your answer to the nearest integer, do not include unit.
Angular velocity of the coil :
[tex]\omega=\dfrac{2\pi n}{60}[/tex]
[tex]\omega=\dfrac{2\times \pi\times 3500}{60}\\\\\omega =366.52\ rad/s[/tex]
Now,
Peak voltage is given by :
[tex]V=NAB\omega\\\\V= 200\times \dfrac{\pi (0.1)^2}{4}\times 0.7\times 366.52\\\\V=403.01\ V[/tex]
Hence, this is the required solution.
What is the potential energy of a 20-kg safe sitting on a shelf 0.5 meters
above the ground?
A-20j
B-10j
C-98j
D-196j
Answer:
C - 98j
Explanation:
Answer:
c- 98j
hope it helped
What would be the luminosity of the Sun if its surface temperature were 4000 K and its radius were 2.0 AU ?
Answer:
0.04558
Explanation:
Luminosity is defined as the total amount of energy a star can produce in a second. Luminosity is given by the formula
σ * A * T⁴
where;
σ = Stefan-Boltzmann constant
= 5.670367 * 10⁻⁸
A= Surface Area
=4πr²
T= Temperature in Kelvin
In the question,
Temperature= 4000K
Radius= 2.0 A.U
Substituting the variables into the equation,
0.000000056704 * 4(3.14)(2.0²) * 4000⁴
= 0.000000056704 * 50.24 * 16000
= 0.04558
The most distant galaxy we have observed is more than
13.2 billion light-years away.
What does this evidence indicate?
e
A. That the universe is less than 13.2 billion years old
B. That the universe will exist for at least 13.2 billion more years
C. That the universe is at least 13.2 billion years old
D. That the universe is infinitely large and has always existed
Answer:
That the universe is at least 13.2 billion years old
Explanation:
The most distant galaxy we have observed is more than 13.2 billion light-years away. this evidence indicates. That the universe is at least 13.2 billion years old
What is the universe?
The universe is the sum total of all existent matter and space. The universe is estimated to be at least 10 billion light-years across and contains an enormous number of galaxies;
It has been expanding since the Big Bang some 13 billion years ago.
The most distant galaxy we have observed is more than 13.2 billion light-years away. this evidence indicates. That the universe is at least 13.2 billion years old
Hence option c is correct.
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Calculate the magnitude of the internal normal force, NB, shear force, VB, and moment, MB, at point B.
Answer:
The answer is "[tex]\bold{MB= \frac{L^2 \ WB}{12}}[/tex]"
Explanation:
please find the complete question in the attached file:
From A to B
[tex]reqAB = \int^{\frac{1}{2}}_{0} \frac{WB}{\frac{L}{2}} x \ dx = \frac{L\ WB}{4}[/tex]
From symmetry [tex]Ay = Cy = FeqAB[/tex]
The FeqAB position is found on a table of common centres:
[tex]\frac{2}{3}( \frac{L}{2})[/tex]
[tex]\to \Sigma_{NB=0} FX = NB=0\\\\\to \Sigma_{VB=0} Fy = \frac{LWB}{4} - \frac{LWB}{4} -VB =0 \\\\\to \Sigma M_{.CCW \ A} = \frac{L}{3} - \frac{LWB}{4} - \frac{L}{2}VB+MB =0 \\\\MB= \frac{L^2 \ WB}{12}[/tex]